What are the solutions to the equation C=x^2-1200x+36400 ?

The solutions to the equation C=x^2-1200x+36400 are x=600+sqrt(C+323600),x=600-sqrt(C+323600)

Find the zeros of C=x^2-1200x+36400

The zeros of C=x^2-1200x+36400 are x=600+sqrt(C+323600),x=600-sqrt(C+323600)

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C = x^{2} - 1200 x + 36400

x = 600 + C + 323600, x = 600 - C + 323600

Solution steps

C = x^{2} - 1200 x + 36400

Switch sides

x^{2} - 1200 x + 36400 = C

Move

C

to the left side

x^{2} - 1200 x + 36400 - C = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1200 ) \pm ( - 1200 ) ^{2} - 4 \cdot 1 \cdot ( 36400 - C )}{2 \cdot 1}

Simplify

(- 1200)^{2} - 4 \cdot 1 \cdot (36400 - C) : 2 C + 323600

x_{1, 2} = \frac{- ( - 1200 ) \pm 2 C + 323600}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1200 ) + 2 C + 323600}{2 \cdot 1}, x_{2} = \frac{- ( - 1200 ) - 2 C + 323600}{2 \cdot 1}

x = \frac{- ( - 1200 ) + 2 C + 323600}{2 \cdot 1} : 600 + C + 323600

x = \frac{- ( - 1200 ) - 2 C + 323600}{2 \cdot 1} : 600 - C + 323600

The solutions to the quadratic equation are:

x = 600 + C + 323600, x = 600 - C + 323600

4 x^{2} - 6 x + 6 = 0

0 = \frac{t ^{2} - t}{3}

8 X^{2} + 2 = 5

2 (x^{2} - 2) = 0

x^{(2)} - 4 x \cdot 3^{(1)} + 27 = 0

What are the solutions to the equation C=x^2-1200x+36400 ?
The solutions to the equation C=x^2-1200x+36400 are x=600+sqrt(C+323600),x=600-sqrt(C+323600)
Find the zeros of C=x^2-1200x+36400
The zeros of C=x^2-1200x+36400 are x=600+sqrt(C+323600),x=600-sqrt(C+323600)